版选修2_3.docx

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1、12.1.2 离散型随机变量的分布列课时目标 1.理解取有限个值的离散型随机变量及其分布列的概念,认识分布列对于刻画随机现象的重要性.2.掌握离散型随机变量分布列的表示方法和性质.3.通过实例(如彩票抽奖),理解二点分布,并能进行简单应用1离散型随机变量的分布列要掌握一个离散型随机变量 X 的取值规律,必须知道:(1)X 所有可能取的值 x1, x2, xn;(2)X 取每一个值 xi的概率 p1, p2, pn.列出表格形式如下:X x1 x2 xi xnP _ _ _ _称这个表为离散型随机变量 X 的_,或称为离散型随机变量 X 的_2离散型随机变量分布列的性质(1)pi_0, i1,2

2、3, n;(2)p1 p2 pn_.3二点分布如果随机变量 X 的分布列为:X 1 0P p q其中 03)_; P(18)_; P(63) P( 4) P( 5) P( 6)0.10.150.20.45; P(18)112 P(X9) P(X10) P(X16)8 , P(6X14) P(X7) P(X8)112 23 P(X14)8 .112 239解 X1,2,3,P(X1) ,C18C2C310 115P(X2) ,C28C12C310 715P(X3) ,C38C02C310 715所以 X 的分布列为X 1 2 3P 115 715 715该考生及格的概率为P(X2) P(X2)

3、 P(X3) .715 715 141510解 因为 X 服从二点分布则 P(X0) , P(X1)1 .C26C211 311 311 811所以 X 的分布列为X 1 0P 811 31111解 由题意及分布列满足的条件知P( 0) P( 1)3 P( 1) P( 1)1,所以 P( 1) ,故 P( 0) .14 34所以 的分布列为7 0 1P 34 1412.解 随机变量 X 取值为 1,2,3,4,5,6.则 P(X1) ;1C16C16 136P(X2) ;3C16C16 336 112P(X3) ;5C16C16 536P(X4) ;7C16C16 736P(X5) ;9C16C16 936 14P(X6) .11C16C16 1136所以两次掷出的最大点数 X 的分布列为X 1 2 3 4 5 6P 136 112 536 736 14 1136

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